\[t\ (ч)\] | \[V(\frac{км}{ч)}\] | \[S\ (км)\] | |
---|---|---|---|
\[I\ авто\] | \[\frac{560}{x + 10}на\ 1\ ч < \searrow\] | \[x + 10\] | \[560\] |
\[\text{II}\ авто\] | \[\frac{560}{x}\] | \[x\] | \[560\] |
\[Составим\ уравнение:\]
\[\frac{560}{x} - \frac{560}{x + 10} = 1\]
\[\frac{560 \cdot (x + 10) - 560x}{x(x + 10)} = 1\]
\[560x + 5600 - 560x = x^{2} + 10x\]
\[x^{2} + 10x - 5600 = 0\]
\[D = b^{2} - 4ac = 100 - 4 \cdot ( - 5600) =\]
\[= 100 + 22\ 400 = 22\ 500\]
\[x_{1} = \frac{- 10 + 150}{2} = \frac{140}{2} =\]
\[= 70\ (\frac{км}{ч) - скорость\ \text{II}\ авто.}\]
\[x_{2} = \frac{- 10 - 150}{2} = \frac{- 160}{2} = - 80 < 0 \Longrightarrow\]
\[\Longrightarrow не\ подходит.\]
\[1)\ 70 + 10 = 80\ (\frac{км}{ч) - V\ I\ авто.}\]
\[Ответ:\ \ V\ І\ авто = 70\ \frac{км}{ч;\ \ \ \ }\]
\[\text{\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ }V\ ІІ\ авто = 80\ \frac{км}{ч.}\]
\[y = - \frac{x - 8}{4} + 1\]
\[y > 0 \Longrightarrow\]
\[\frac{- (x - 8)}{4} + 1 > 0\]
\[\frac{- x + 8}{4} > - 1\]
\[- x + 8 > - 4\]
\[- x > - 12\]
\[x < 12\]
\[Ответ:при\ \ x \in ( - \infty;12).\]
\[\left\{ \begin{matrix} 5 \cdot (2x - 1) - 3 \cdot (3x + 6) < 2 \\ 2x - 17 > 0\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ \ }\]
\[\left\{ \begin{matrix} 10x - 5 - 9x - 18 < 2 \\ 2x > 17\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \]
\[\left\{ \begin{matrix} x < 25\ \\ x > 8,5 \\ \end{matrix} \right.\ \]
\[Ответ:\ x \in (8,5;25).\]
\[\left( \sqrt{10} + \sqrt{5} \right) \cdot \sqrt{20} - 5\sqrt{8} =\]
\[= 10\sqrt{2} + 10 - 10\sqrt{2} = 10.\]
\[\left( \frac{2}{x^{2} - 4} + \frac{1}{2x - x^{2}} \right)\ :\frac{1}{x^{2} + 4x + 4} =\]
\[= \left( \frac{2}{(x - 2)(x + 2)} + \frac{1}{x(2 - x)} \right)\ :\frac{1}{(x + 2)^{2}} =\]
\[= \left( \frac{2x - (x + 2)}{x(x - 2)(x + 2)} \right) \cdot (x + 2)^{2} =\]
\[= \frac{2x - x - 2}{x(x - 2)(x + 2)} \cdot (x + 2)^{2} =\]
\[= \frac{(x - 2)(x + 2)²}{x(x - 2)(x + 2)} = \frac{x + 2}{x}\]